Question

I want to know if this represents a tail-recursion. And if it isn’t how can I do it.

 countP :: [a] -> (a->Bool) -> Int
 countP [] _ = 0
 countP (x:xs) p = countP_aux (x:xs) p 0

 countP_aux [] _ _ = 0
 countP_aux (x:xs) p acumul
                        |p x==True = (countP_aux xs p (acumul))+1
                        |otherwise = (countP_aux xs p (acumul))

  countP [1,2,3] (>0)
  3
  (72 reductions, 95 cells)

This exercise show how many values in a list are verified by p condition.
Thanks

Answer 1

This is not tail recursive because of

(countP_aux xs p (acumul))+1

Tail calls should return the result of the recursive call, rather than doing calculation with the result of the recursive call.

You can convert a non-tail recursive function to be tail-recursive by using an accumulator where you perform the additional work, i.e.

Say you have a simple counting function

f a
  | a < 1 = 0 
  | otherwise = f (a-1) + 1

You can make it tail recursive like so:

f' acc a = 
  | a < 1 = acc 
  | otherwise = f' (acc + 1) (a-1)
f = f' 0